3 Answers
3

The notation is a little neater if we do the induction step from $n$ to $n+1$ instead of from $n-1$ to $n$. My induction hypothesis is that for all $x$ and $y$, $$(x+y)^{\overline n} = \sum\limits_{k=0}^n \binom{n}{k}x^{\overline k}y^{\overline{n-k}}.$$

I want to show that for all $x$ and $y$, $$(x+y)^{\overline {n+1}} = \sum\limits_{k=0}^{n+1}\binom{n+1}{k}x^{\overline k}y^{\overline{n+1-k}}.$$

I’ll be using the fact that $u^{\overline {m+1}} = u(u+1)^{\overline m}$.

At step $(1)$ I split off the last term of the sum. At step $(2)$ I used the basic binomial recursion. At step $(3)$ I did an index shift on the first sum: the $k=0$ term is $0$, so $k$ might as well run from $1$ to $n$ and can then be replaced by $k-1$, which runs from $0$ to $n-1$. At step $(4)$ I combined the separated term from step $(1)$ with the first summation. At step $(5)$ I used the fact mentioned just before the computation, and at step $(6)$ I applied the induction hypothesis. The rest is just algebra and again the fact mentioned before the computation.

Interesting way of prooving this, however i have to do this via induction. Still thanks for sharing the idea.
–
Christian IvicevicSep 17 '11 at 9:35

1

@Christian: I knew you had to use induction. Sometimes other people reading the question are interested in different ways to prove the result, though, and I posted this proof for them as well.
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Mike SpiveySep 17 '11 at 12:53